Question

The following notation is used in the problems: \(M=\) mass, \(\bar{x}, \bar{y}, \bar{z}=\) coordinates of center of mass (or centroid if the density is constant), \(I=\) moment of inertia (about axis stated), \(I_{x}, I_{y}, I_{z}=\) moments of inertia about \(x, y, z\) axes, \(I_{m}=\) moment of inertia (about axis stated) through the center of mass. Note: It is customary to give answers for \(I, I_{m}, I_{x},\) etc., as multiples of \(M\) (for example, \(I=\frac{1}{3} M l^{2}\) ). A rectangular lamina has vertices (0,0),(0,2),(3,0),(3,2) and density \(x y .\) Find (a) \(M\), (b) \(\bar{x}, \bar{y}\), (c) \(I_{x}, I_{y}\), (d) \( I_{m}\) about an axis parallel to the \(z\) axis. Hint: Use the parallel axis theorem and the perpendicular axis theorem.

Short answer

a) \( M = 9 \), b) \( \bar{x} = 2 \), \( \bar{y} = \frac{4}{3} \), c) \( I_x = 18 \), \( I_y = 40.5 \), d) \( I_m = 6.5 \).

Step-by-step solution

- Calculate Mass (M)

To find the mass of the rectangular lamina with vertices at (0,0), (0,2), (3,0), (3,2) and density function \( xy \), integrate the density over the area of the rectangle:\[ M = \int_0^3 \int_0^2 x y \ dy \ dx \].

- Evaluate Inner Integral for M

First, integrate with respect to y:\[ \int_0^2 x y \ dy = x \left[ \frac{y^2}{2} \right]_0^2 = x \cdot 2 = 2x \].

- Evaluate Outer Integral for M

Next, integrate the result with respect to x:\[ M = \int_0^3 2 x \ dx = 2 \left[ \frac{x^2}{2} \right]_0^3 = \left[ x^2 \right]_0^3 = 3^2 = 9 \].So, \( M = 9 \).

- Calculate \( \bar{x} \) and \( \bar{y} \)

The coordinates of the center of mass are given by:\[ \bar{x} = \frac{1}{M} \int_0^3 \int_0^2 x^2 y \ dy \ dx \] and \[ \bar{y} = \frac{1}{M} \int_0^3 \int_0^2 x y^2 \ dy \ dx \].

- Evaluate Inner Integrals for Center of Mass

First integrate \( \int_0^2 x^2 y \ dy \):\[ \int_0^2 x^2 y \ dy = x^2 \left[ \frac{y^2}{2} \right]_0^2 = x^2 \cdot 2 = 2x^2 \].Now integrate \( \int_0^2 x y^2 \ dy \):\[ \int_0^2 x y^2 \ dy = x \left[ \frac{y^3}{3} \right]_0^2 = x \cdot \frac{8}{3} = \frac{8x}{3} \].

- Evaluate Outer Integrals for Center of Mass

Now integrate the results with respect to x:\[ \bar{x} = \frac{1}{9} \int_0^3 2 x^2 \ dx = \frac{1}{9} \cdot 2 \left[ \frac{x^3}{3} \right]_0^3 = \frac{1}{9} \cdot 2 \cdot \frac{27}{3} = 2 \].\[ \bar{y} = \frac{1}{9} \int_0^3 \frac{8 x}{3} \ dx = \frac{1}{9} \cdot \frac{8}{3} \left[ \frac{x^2}{2} \right]_0^3 = \frac{1}{9} \cdot \frac{8}{3} \cdot \frac{9}{2} = \frac{4}{3} \].So, \( \bar{x} = 2 \) and \( \bar{y} = \frac{4}{3} \).

- Calculate \( I_x \) and \( I_y \)

The moments of inertia about the x and y axes are given by:\[ I_x = \int_0^3 \int_0^2 x y^3 \ dy \ dx \] and \[ I_y = \int_0^3 \int_0^2 x^3 y \ dy \ dx \].

- Evaluate Inner Integrals for \( I_x \) and \( I_y \)

First integrate \( \int_0^2 x y^3 \ dy \):\[ \int_0^2 x y^3 \ dy = x \left[ \frac{y^4}{4} \right]_0^2 = x \cdot 4 = 4x \].Now integrate \( \int_0^2 x^3 y \ dy \):\[ \int_0^2 x^3 y \ dy = x^3 \left[ \frac{y^2}{2} \right]_0^2 = x^3 \cdot 2 = 2x^3 \].

- Evaluate Outer Integrals for \( I_x \) and \( I_y \)

Now integrate the results with respect to x:\[ I_x = \int_0^3 4 x \ dx = 4 \left[ \frac{x^2}{2} \right]_0^3 = 4 \cdot \frac{9}{2} = 18 \].\[ I_y = \int_0^3 2 x^3 \ dx = 2 \left[ \frac{x^4}{4} \right]_0^3 = 2 \cdot \frac{81}{4} = 40.5 \].So, \( I_x = 18 \) and \( I_y = 40.5 \).

- Use the Perpendicular Axis Theorem

The perpendicular axis theorem states that for a lamina in the xy-plane, \[ I_z = I_x + I_y \].So, \( I_z = 18 + 40.5 = 58.5 \).

- Calculate \( I_m \) Using Parallel Axis Theorem

The parallel axis theorem states:\[ I_m = I_z - M \left(\bar{x}^2 + \bar{y}^2\right) \].Substitute the known values:\[ I_m = 58.5 - 9 \left(2^2 + (\frac{4}{3})^2\right) = 58.5 - 9 \left(4 + \frac{16}{9}\right) = 58.5 - 9 \cdot \frac{52}{9} = 58.5 - 52 = 6.5 \].

Key concepts

center of mass

The center of mass is the point where the entire mass of an object can be thought to be concentrated. For a rectangular lamina with variable density, the center of mass (or centroid) coordinates, \(\bar{x}\) and \(\bar{y}\), can be found using integration. To determine these coordinates, we divide the moments (weighted averages) by the total mass M of the lamina.

• The coordinate \(\bar{x}\) is given by: \(\bar{x} = \frac{1}{M} \int_0^3 \int_0^2 x^2 y \ dy \ dx\)


• The coordinate \(\bar{y}\) is given by: \(\bar{y} = \frac{1}{M} \int_0^3 \int_0^2 x y^2 \ dy \ dx\)

In these integrals, the density function \(xy\) is incorporated, effectively weighing each part of the lamina according to its density. This approach ensures accurate location of the center of mass in objects with non-uniform density.

perpendicular axis theorem

The perpendicular axis theorem provides a relationship between the moments of inertia about three mutually perpendicular axes. For a plane lamina (a flat object) lying in the xy-plane, the theorem states:

\[ I_z = I_x + I_y \]

Here:

• \( I_z \) is the moment of inertia about the z-axis, which is perpendicular to the lamina


• \( I_x \) and \( I_y \) are the moments of inertia about the x and y axes, respectively

By summing the moments of inertia along the x and y axes, we obtain the moment of inertia along the z-axis. This principle greatly simplifies the calculation and helps in determining the rotational properties of the object.

parallel axis theorem

The parallel axis theorem is used to find the moment of inertia of an object about any axis, given the moment of inertia about a parallel axis through the center of mass. It states:

\[ I = I_m + M \cdot d^2 \]

Here:

• \( I \) is the moment of inertia about the new axis


• \( I_m \) is the moment of inertia about the axis through the center of mass


• \( M \) is the mass of the object


• \( d \) is the distance between the center of mass and the new axis

This theorem is particularly useful when dealing with complex shapes or when calculating moments of inertia about multiple axes. For our problem, the parallel axis theorem helps in determining \( I_m \) from \( I_z \), integrating the mass and the square of the distance from the center of mass.

density function integration

Density function integration is a pivotal concept in physics and engineering for calculating mass, center of mass, and moments of inertia, especially for objects with non-uniform density. For a given density function \( \rho(x, y) \), integration over the object's area involves:

1. **Total Mass (M)**: This is found by integrating the density function over the entire region:

\[ M = \int \int \rho \ dA \]

2. **Center of Mass Coordinates**: These coordinates are weighted averages that locate the effective center:

• \( \bar{x} = \frac{1}{M} \int \int x \rho \ dA \)


• \( \bar{y} = \frac{1}{M} \int \int y \rho \ dA \)

3. **Moments of Inertia**: These are calculated by integrating density multiplied by the square of the distance from the axis:

• \( I_x = \int \int y^2 \rho \ dA \)
• \( I_y = \int \int x^2 \rho \ dA \)

In our problem, the integration of the density function \( xy \) across the rectangular lamina helps in computing all relevant physical properties, confirming the critical role of density function integration in mechanics.